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I can do the following derivation using solid angle:

$$\unicode{x222F} \dfrac{\hat{r} \cdot \vec{dS}}{r^2} =\unicode{x222F} \dfrac{dS \cos\alpha}{r^2} =\int^{2\pi}_0 \int^\pi_0 \sin\theta\ d\theta\ d\phi=4\pi$$

However I have no clue how to show that:

$$\unicode{x222F} \dfrac{\hat{r} \times \vec{dS}}{r^2}=0$$

Please give some hint to solve this problem.

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    $\begingroup$ You can use the LaTeX code \oiint for $\oiint$. EDIT: Apparently not. One moment $\endgroup$ – Akiva Weinberger Nov 28 '18 at 6:53
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    $\begingroup$ Wow, apparently Unicode is the only way to do it. That's annoying. $\endgroup$ – Akiva Weinberger Nov 28 '18 at 6:58
  • $\begingroup$ What is the domain of integration? Any closed surface or simply a sphere centered at the origin? $\endgroup$ – Robert Z Nov 28 '18 at 7:09
  • $\begingroup$ Any closed surface $\endgroup$ – Joe Nov 28 '18 at 7:10
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By Prove $ \oint_{\partial V} (\mathbf{\hat{n}} \times \mathbf{A}) \; \mathrm{d}S = \int_V (\nabla \times \mathbf{A}) \; \mathrm{d}V $ (see also wiki-link) $$\oint_S \dfrac{\hat{r} \times \vec{dS}}{r^2}=\oint_S \left(\dfrac{\hat{r} }{r^2}\times \hat{n}\right) dS=-\int_V\left(\nabla \times\left(\dfrac{\hat{r} }{r^2}\right)\right)dV=0$$ where at the last step we used the fact that $$\dfrac{\hat{r}}{r^2}=\nabla\left(\frac{1}{r}\right)$$ and the curl of a gradient is identically zero, i.e. $\nabla \times \nabla (1/r)=0$.

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Just "dot" the integral with constant vector $\vec{k}$ and apply divergence theorem.

Let $V$ be the volume bounded by closed surface $S$.

Let us first consider the case $V$ doesn't contain origin. We have

$$\vec{k} \cdot \int_{S} \frac{\hat{r} \times \vec{dS}}{r^2} = \int_{S} \frac{\vec{k} \times \hat{r} }{r^2} \cdot \vec{dS} = \int_V \nabla \cdot \frac{\vec{k} \times \hat{r} }{r^2} dV = -\int_V \vec{k}\cdot \left(\nabla \times \frac{\hat{r}}{r^2}\right)dV\\ = \vec{k} \cdot \int_{V} \nabla \times \nabla \frac{1}{r} dV = \vec{k} \cdot \int_{V} \vec{0} dV = \vec{k}\cdot \vec{0}$$

Since this is true for all $\vec{k}$, we get $$\int_{S} \frac{\hat{r} \times \vec{dS}}{r^2} = \vec{0}$$ When $V$ contains origin, apply above result to volume $V \setminus B(0,\epsilon)$ where $B(0,\epsilon)$ is a small ball centered at origin with radius $\epsilon$ which lies completely inside $V$. We obtain

$$\int_{\partial( V \setminus B(0,\epsilon))}\frac{\hat{r} \times \vec{dS}}{r^2} = \vec{0} \implies \int_{S} \frac{\hat{r} \times \vec{dS}}{r^2} = \int_{|r|=\epsilon} \frac{\hat{r} \times \vec{dS}}{r^2} $$

On the sphere $|r| = \epsilon$, $\hat{r}$ and $\vec{dS}$ is pointing towards same direction. This means $\displaystyle\;\frac{\hat{r} \times \vec{dS}}{r^2} = \vec{0}$ there and the integral on RHS vanishes.

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